Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $n \neq 0$. $q = \dfrac{n - 1}{n^2 - 10n + 9} \div \dfrac{2n - 8}{-8n + 72} $
Answer: Dividing by an expression is the same as multiplying by its inverse. $q = \dfrac{n - 1}{n^2 - 10n + 9} \times \dfrac{-8n + 72}{2n - 8} $ First factor the quadratic. $q = \dfrac{n - 1}{(n - 9)(n - 1)} \times \dfrac{-8n + 72}{2n - 8} $ Then factor out any other terms. $q = \dfrac{n - 1}{(n - 9)(n - 1)} \times \dfrac{-8(n - 9)}{2(n - 4)} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac{ (n - 1) \times -8(n - 9) } { (n - 9)(n - 1) \times 2(n - 4) } $ $q = \dfrac{ -8(n - 1)(n - 9)}{ 2(n - 9)(n - 1)(n - 4)} $ Notice that $(n - 1)$ and $(n - 9)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac{ -8(n - 1)\cancel{(n - 9)}}{ 2\cancel{(n - 9)}(n - 1)(n - 4)} $ We are dividing by $n - 9$ , so $n - 9 \neq 0$ Therefore, $n \neq 9$ $q = \dfrac{ -8\cancel{(n - 1)}\cancel{(n - 9)}}{ 2\cancel{(n - 9)}\cancel{(n - 1)}(n - 4)} $ We are dividing by $n - 1$ , so $n - 1 \neq 0$ Therefore, $n \neq 1$ $q = \dfrac{-8}{2(n - 4)} $ $q = \dfrac{-4}{n - 4} ; \space n \neq 9 ; \space n \neq 1 $